## Precalculus (6th Edition)

$\color{blue}{(D' \cap U) \cup E = \left\{3, 4, 5, 6, 7, 8, 9, 10\right\}}$
RECALL: (1) $\cup$ represents the union of sets, which is the set that contains the combined elements of the sets involved. (2) $\cap$ represents the intersection of sets, which is the set that contains the elements common to the sets involved. (3) $A'$ represents the complement of $A$, which is the set that contains the elements of the universal set $U$ that are not in $A$> Given: $D=\left\{1, 2, 3\right\}$ $E=\left\{3, 7\right\}$ $U=\left\{1, 2, 3, 4, 5, 6, 7, 8, 9, 105\right\}$ Thus, $D'=\left\{4, 5, 6, 7, 8,9, 10\right\}$ The intersection of the universal set and any of its subset is the subset itself. Thus, $D'\cap U=D'=\left\{4, 5, 6, 7, 8, 9, 10\right\}$ Combining the elements of the set above and the set $E$ gives: $\color{blue}{(D' \cap U) \cup E = \left\{3, 4, 5, 6, 7, 8, 9, 10\right\}}$